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In the last video we figured
out that if we had a proton

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coming into the right at a
velocity of 6 times 10 to the

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seventh meters per second.

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So the magnitude of the
velocity is 1/5

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the speed of light.

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And if it were to cross this
magnetic field, we used this

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formula to figure out that the
magnitude of the force on this

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proton would be 4.8 times 10
to the negative 12 newtons.

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And then the direction, we
used our right hand rule

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because it's a cross product.

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And we figured out that it would
be perpendicular-- well,

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it has to be perpendicular to
both, because we're taking the

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cross product-- and right when
it enters, the net force will

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be downwards.

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But then think about
what happens.

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If you have a downward force
right there, then the particle

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will be deflected downward a
little bit, so its velocity

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vector will then look
something like that.

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But it's still in the magnetic
field, right?

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And not only is it still in the
magnetic field, but since

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the particle is still moving
within the plane of your video

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screen, it's still completely
perpendicular to

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the magnetic field.

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And so the magnitude of the
force on the moving particle

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won't change, just the
direction will.

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Because if we do the right hand
rule here, but if we just

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move our hand down a little bit,
if we tilt it down, then

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our thumb's going to be pointing
in this direction.

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And that just keeps happening.

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It gets deflected that
way a little bit.

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So the magnitude of the velocity
doesn't ever change.

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It always stays perpendicular to
the magnetic field because

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it's always staying
in this plane.

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But the orientation does change
within the plane.

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And because of that, because
the orientation of the

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velocity changes,
the orientation

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of the force changes.

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So when the velocity is here,
the force is perpendicular.

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So it acts as kind of a
centripetal force, and so the

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particle will start moving
in a circle.

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So let's see if we can break
out our toolkit from what

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we've learned before in
classical mechanics, and

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figure out what the radius
of that circle is.

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And that might seem more
daunting than it really is.

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Well, what do we know about
centripetal forces and

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radiuses of circles,
et cetera?

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So, what is the formula
for centripetal force?

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And we proved it many, many
videos ago, early in the

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physics playlist. Well,
centripetal acceleration is

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the magnitude of the velocity
vector squared over the radius

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of the circle.

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And since this is acceleration,
if we want to

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know the centripetal force,
it's just the mass times

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acceleration.

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So it's the mass of the
particle, or the object in

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question, times the magnitude
of its velocity squared

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divided by the radius
of the circle.

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In this case, this is the
radius of the circle.

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And that's what we're going
to try to solve for.

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And what do we know about
the centripetal force?

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What is causing the
centripetal force?

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Well, it's the magnetic field
and we've figured that out.

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This is going to be equal to
this, which we figured out is

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going to be equal to-- at least
the magnitudes-- the

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magnitude of this is equal
to the magnitude of this.

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And that magnitude is
4.8 times 10 to

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the minus 12 newtons.

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And so the radius is going to
be-- let's see, if we flip

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both sides of this equation,
we get radius over mass

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velocity squared is equal
to 1 over 4.8 times 10

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to the minus 12.

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I could just figure out what
that number is, but I won't

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worry about that now.

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Then we can multiply both sides
times this mv squared.

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And we get that the radius of
the circle is going to be

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equal to the mass of the proton
times the magnitude of

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its velocity squared divided
by the force from

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the magnetic field.

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The centripetal force.

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4.8 times 10 to the
minus 12 newtons.

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And the radius should be in
meters, since everything is

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kind of in the standard
SI units.

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And let's see if we can
figure this out.

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Get our calculator.

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And this is where that constant
function is useful

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again, because what is
the mass of a proton?

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Well, that's something that I

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personally don't have memorized.

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But if we go into the built-in
constants on the

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TI-85-- let's see more.

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Mass of a proton.

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This is mass of an electron.

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This is mass of a proton.

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So mass of a proton-- that's
what we care about-- times the

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magnitude of the velocity
squared.

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What was the velocity?

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It was 6 times 10 to the seventh
meters per second.

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So times 6 times 10
to the seventh

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meters per second squared.

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And all of that divided
by the magnitude of

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the centripetal force.

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Which is the force that's
being generated by the

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magnetic field.

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That's 4.8 times 10 to
the negative 12.

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Divided by 4.8 E minus 12.

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Let's see.

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Hopefully we don't get
something funky.

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There we go.

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That's actually a pretty
neat number.

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1.25 meters.

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That's actually a number
that we can imagine.

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So if you have a proton going
in this direction at 1/5 the

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speed of light through a--
what'd I say it was?

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It was a 0.5 tesla magnetic
field, where the vectors are

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pointing out of the video.

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We have just shown that this
proton will go in a circle of

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radius 1.25 meters.

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Which is neat because
it's a number that I

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can actually visualize.

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And so this whole business of
magnetic fields making charged

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particles go into circles, this
is one of the few times

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that I can actually say has a
direct application into things

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that you've seen.

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Namely, your TV.

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Or at least the old-school TVs.
The non-plasma or LCD

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TVs, your cathode ray TVs,
take advantage of this.

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Where you essentially
have a beam of

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not protons but electrons.

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And a magnet-- if you take
apart a TV, which I don't

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think you should do, because
you're more likely to hurt

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yourself because there's a
vacuum in there that can

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implode, and all that-- but
essentially, you have a magnet

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that deflects this electron beam
and does it really fast

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so it scans your entire screen
of different intensities, and

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that's what forms the image.

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I won't go into that detail.

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Maybe one day I'll do a whole
video on how TVs work.

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So that's one application of a
magnetic field causing a beam

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of charged particles to curve.

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And then the other application,
and this is

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actually one where it's actually
useful to make the

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particle go in a circle, is
these cyclotrons that you read

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about, where they take these
protons and they make them go

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in circles really, really
fast, and then

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they smash them together.

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Well, have you ever wondered,
how do they even make a proton

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go in a circle?

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It's not like you could
hold it and guide

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it around in a circle.

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Well, that's what they do.

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They pass it through an
appropriate strength magnetic

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field, and it curves the path
of the proton so that it can

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keep going through the same
field over and over again.

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And then they can actually use
those electric fields.

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I don't claim to have any
expertise in this, but then

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they can keep speeding it up
using the same devices,

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because it keeps passing
through the

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same part of the collider.

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And then once it collides,
you've

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probably seen those pictures.

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You know, that you spend
billions of dollars on

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supercolliders, and you end
up with these pictures.

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And somehow these physicists
are able to take these

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pictures and say, oh, this is
some new particle because of

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the way it moved.

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Well, what they're actually
talking about is these are

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moving at relativistic speeds.

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And since they're at
relativistic speeds as they

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move at different velocities,
their mass is

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changing, and all that.

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But the basic idea is what
we just learned.

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They move in circles.

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They move in circles because
they're going through a

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magnetic field.

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But their radiuses are different
because their

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charges and their velocities
are going to be different.

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And actually some will move
to the left and some

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will move to the right.

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And that might be because
they're positive or negative,

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and then the radius will be
dependent on their masses.

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Anyway, I don't want
to confuse you.

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But I just wanted to show you
that we actually are touching

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on some physics that
a physicist would

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actually care about.

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Now with that said, what would
have happened if this wasn't a

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proton but if this was an
electron moving at this

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velocity at 6 times 10 to the
seventh meters per second

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through a 0.5 tesla
magnetic field

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popping out of this video.

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What would have happened?

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Well, this formula would
have still been safe.

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The magnitude of the force is
the charge-- but it wouldn't

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have been the charge of a
proton, it would have been the

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00:09:02,630 --> 00:09:06,040
charge of an electron, times 6
times 10 to the seventh meters

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00:09:06,040 --> 00:09:08,490
per second times 0.5 teslas.

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So what's the difference between
the charge of a proton

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and the charge of an electron?

202
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Well, the charge of an
electron is negative.

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So if this was an electron,
then the net force would

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actually end up being
a negative number.

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So what does that mean?

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Well, when we used the right
hand rule with the proton

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example, we said that the-- at
least when the proton is

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moving in this direction--
that the net

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force would be downwards.

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00:09:30,690 --> 00:09:33,880
But now, all of a sudden, if we
reverse the charge, if we

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say we have a negative charge--
the same magnitude

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00:09:35,850 --> 00:09:36,980
but it's negative,
because it's an

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00:09:36,980 --> 00:09:39,410
electron-- what happens?

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The force is now in this
direction, using the right

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00:09:41,910 --> 00:09:43,880
hand rule, but it
is a negative.

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So really it's going to be a
positive force of the same

217
00:09:46,530 --> 00:09:48,640
magnitude in this direction.

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00:09:48,640 --> 00:09:50,700
So if we have a proton,
it'll go in a

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circle in this direction.

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It'll go like this.

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But if we have an electron,
it'll go in a circle of the

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other direction.

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Now let me ask a question.

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Is that circle going
to be a tighter

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circle or a wider circle?

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Well, the mass of an electron is
a lot smaller than the mass

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of a proton.

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And we had the radius is equal
to the mass times the velocity

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squared divided by the
centripetal force.

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So this mass is smaller
and the radius

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is going to be smaller.

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So the electron's path would
actually move up and it would

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be a smaller radius.

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Actually proportional to the
difference in their radiuses

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is the difference in their
masses, actually.

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But that would be the path
of the electron.

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Anyway, I thought you'd be
interested in that, as well.

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I have run out of time.

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I will see you in
the next video.